Predict the molecular structure for each of the following. (See Exercises 105 and \(106 .\) ) a. \(\mathrm{BrFI}_{2}\) b. \(\mathrm{XeO}_{2} \mathrm{~F}_{2}\) c. \(\mathrm{TeF}_{2} \mathrm{Cl}_{3}^{-}\) For each formula there are at least two different structures that can be drawn using the same central atom. Draw all possible structures for each formula.

First, we need to determine the central atom, which is the least electronegative atom (excluding Hydrogen). In BrFI2, the central atom is Br. Now, let's count the number of electron domains, including bonding pairs and lone pairs. Br has 7 valence electrons, and each of the F and I atoms contributes 1 electron for bonding. Therefore, we have a total of 9 electrons to distribute. That means there are 4 electron domains (3 bonding pairs and 1 lone pair) around the Br atom. Using the VSEPR model, the electron geometry for 4 electron domains is tetrahedral. However, since it is specified that there are at least two different structures that can be drawn, we can arrange the F and I atoms differently: - Structure 1: Br with two F atoms and one I atom bonded, with a lone pair. - Structure 2: Br with two I atoms and one F atom bonded, with a lone pair.

The central atom for XeO2F2 is Xe. Now, let's count the number of electron domains. Xe has 8 valence electrons, and each of the O and F atoms contributes 1 electron for bonding. Therefore, we have a total of 12 electrons to distribute. That means there are 5 electron domains (4 bonding pairs and 1 lone pair) around the Xe atom. Using the VSEPR model, the electron geometry for 5 electron domains is trigonal bipyramidal. We should consider two different arrangements of O and F atoms: - Structure 1: Two O atoms occupy the axial positions and two F atoms occupy the equatorial positions. - Structure 2: Two F atoms occupy the axial positions and two O atoms occupy the equatorial positions.

The central atom for TeF2Cl3- is Te. Now, let's count the number of electron domains. Te has 6 valence electrons and as an anion, it has one more added electron to its outer shell, making it 7. Each of the F and Cl atoms contributes 1 electron for bonding. Therefore, we have a total of 12 electrons to distribute. That means there are 6 electron domains (5 bonding pairs and 1 lone pair) around the Te atom. Using the VSEPR model, the electron geometry for 6 electron domains is octahedral. We should consider two different arrangements of F and Cl atoms: - Structure 1: One F and three Cl atoms occupy the equatorial plane while another F atom occupies the axial position. - Structure 2: Two F atoms and two Cl atoms occupy the equatorial plane while another Cl atom occupies the axial position. Now, you can depict the structures in two-dimensional Lewis structures or three-dimensional molecular geometry models.